A) \[9.66\text{ }m\]
B) \[7.89\text{ }m\]
C) \[8.66\text{ }m\]
D) \[7.64\text{ }m\]
Correct Answer: C
Solution :
Suppose height of the tower AB = x m and distance BC = ym. ln rt. \[\Delta \,ABC,\,\,\frac{AB}{BC}=\tan {{60}^{o}}\] \[\Rightarrow \] \[\frac{x}{y}=\sqrt{3}\Rightarrow y=\frac{x}{\sqrt{3}}\] In rt. \[\Delta \,ABC,\frac{AB}{BC}=\tan {{60}^{o}}\] \[\Rightarrow \]\[\frac{x}{y}=\sqrt{3}\Rightarrow y=\frac{x}{\sqrt{3}}\] In rt, \[\Delta \,ABD,\,\,\frac{AB}{DB}=\tan {{30}^{o}}\] \[\Rightarrow \] \[\frac{x}{10+y}=\frac{1}{\sqrt{3}}\Rightarrow \sqrt{3}x=10+y\] ?.(2) Putting the value of y in (2), we get \[\sqrt{3}x=10+\frac{x}{\sqrt{3}}\]You need to login to perform this action.
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